Chapter 4 Managerial Statistics solutions

8/13/2019 Chapter 4 Managerial Statistics solutions

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Chapter 4: Probability 1

Chapter 4

Probability

LEARNING OBJECTIVES

The main objective of Chapter 4 is to help you understand the basic principles of

probability, specifically enabling you to

1. Comprehend the different ways of assigning probability.

2. Understand and apply marginal, union, joint, and conditional probabilities.

3. Select the appropriate law of probability to use in solving problems.

4. Solve problems using the laws of probability, including the law of addition, the

law of multiplication , and the law of conditional probability.

5. Revise probabilities using Bayes' rule.

CHAPTER TEACHING STRATEGY

Students can be motivated to study probability by realizing that the field of

probability has some stand-alone application in their lives in such applied areas human

resource analysis, actuarial science, and gaming. In addition, students should understandthat much of the rest of the course is based on probabilities even though they will not be

directly applying many of these formulas in other chapters.

This chapter is frustrating for the learner because probability problems can be

approached by using several different techniques. Whereas, in many chapters of this text,students will approach problems by using one standard technique, in chapter 4, different

students will often use different approaches to the same problem. The text attempts to

emphasize this point and underscore it by presenting several different ways to solve

probability problems. The probability rules and laws presented in the chapter can

virtually always be used in solving probability problems. However, it is sometimes easier

to construct a probability matrix or a tree diagram or use the sample space to solve the

problem. If the student is aware that what they have at their hands is an array of tools or

techniques, they will be less overwhelmed in approaching a probability problem. An


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attempt has been made to differentiate the several types of probabilities so that students

can sort out the various types of problems.

In teaching students how to construct a probability matrix, emphasize that it isusually best to place only one variable along each of the two dimensions of the matrix.

(That is place Mastercard with yes/no on one axis and Visa with yes/no on the other

instead of trying to place Mastercard and Visa along the same axis).

This particular chapter is very amenable to the use of visual aids. Students enjoy

rolling dice, tossing coins, and drawing cards as a part of the class experience.

Of all the chapters in the book, it is most imperative that students work a lot ofproblems in this chapter. Probability problems are so varied and individualized that a

significant portion of the learning comes in the doing. Experience is an important factor

in working probability problems.

Section 4.8 on Bayes theorem can be skipped in a one-semester course withoutlosing any continuity. This section is a prerequisite to the chapter 18 presentation of

revising probabilities in light of sample information (section 18.4).

CHAPTER OUTLINE

4.1 Introduction to Probability4.2 Methods of Assigning Probabilities

Classical Method of Assigning Probabilities

Relative Frequency of Occurrence

Subjective Probability

4.3 Structure of Probability

ExperimentEvent

Elementary Events

Sample Space

Unions and Intersections

Mutually Exclusive EventsIndependent Events

Collectively Exhaustive Events

Complimentary Events

Counting the Possibilities

The mn Counting Rule

Sampling from a Population with Replacement

Combinations: Sampling from a Population Without Replacement

4.4 Marginal, Union, Joint, and Conditional Probabilities


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4.5 Addition Laws

Probability MatricesComplement of a Union

Special Law of Addition

4.6 Multiplication Laws

General Law of Multiplication

Special Law of Multiplication

4.7 Conditional Probability

Independent Events

4.8 Revision of Probabilities: Bayes' Rule

KEY TERMS

A Priori IntersectionBayes' Rule Joint Probability

Classical Method of Assigning Probabilities Marginal Probability

Collectively Exhaustive Events mnCounting RuleCombinations Mutually Exclusive Events

Complement of a Union Probability Matrix

Complementary Events Relative Frequency of Occurrence

Conditional Probability Sample Space

Elementary Events Set NotationEvent Subjective Probability

Experiment Union

Independent Events Union Probability

SOLUTIONS TO PROBLEMS IN CHAPTER 4

4.1 Enumeration of the six parts: D1, D2, D3, A4, A5, A6D = Defective part

A = Acceptable part

Sample Space:

D1D2, D2D3, D3A5

D1D3, D2A4, D3A6D1A4, D2A5, A4A5

D1A5, D2A6, A4A6


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D1A6, D3A4, A5A6

There are 15 members of the sample space

The probability of selecting exactly one defect out of

two is:

9/15 = .60

4.2 X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9}

and Z = {1, 2, 3, 4, 7,}

a) X Z = {1, 2, 3, 4, 5, 7, 8, 9}b) X_ Y = {7, 9}c) X_ Z = {1, 3, 7}d) X Y Z = {1, 2, 3, 4, 5, 7, 8, 9}e) X_ Y_ Z = {7}f) (X Y)_Z = {1, 2, 3, 4, 5, 7, 8, 9}_ {1, 2, 3, 4, 7} = {1, 2, 3, 4, 7}g) (Y_ Z) (X_ Y) = {2, 4, 7} {7, 9} ={2, 4, 7, 9}h) X or Y = X Y = {1, 2, 3, 4, 5, 7, 8, 9}i) Y and Z = Y_ Z = {2, 4, 7}

4.3 If A = {2, 6, 12, 24} and the population is the positive even numbers through 30,

A = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}

4.4 6(4)(3)(3) = 216

4.5 Enumeration of the six parts: D1, D2, A1, A2, A3, A4

D = Defective partA = Acceptable part

Sample Space:

D1D2A1, D1D2A2, D1D2A3,

D1D2A4, D1A1A2, D1A1A3,D1A1A4, D1A2A3, D1A2A4,

D1A3A4, D2A1A2, D2A1A3,D2A1A4, D2A2A3, D2A2A4,

D2A3A4, A1A2A3, A1A2A4,

A1A3A4, A2A3A4

Combinations are used to counting the sample space because sampling is done

without replacement.


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6C3=!3!3

!6 = 20

Probability that one of three is defective is:

12/20 = 3/5 .60

There are 20 members of the sample space and 12 of them have 1 defective part.

4.6 107= 10,000,000 different numbers

4.7 20C6=

!14!6

!20= 38,760

It is assumed here that 6 different (without replacement) employees are to be

selected.

4.8 P(A) = .10, P(B) = .12, P(C) = .21

P(A_ C) = .05 P(B C) = .03

a) P(A C) = P(A) + P(C) - P(A_ C) = .10 + .21 - .05 = .26

b) P(B C) = P(B) + P(C) - P(B_C) = .12 + .21 - .03 = .30

c) If A, B mutually exclusive, P(A B) = P(A) + P(B) = .10 + .12 = .22

4.9

D E F

A 5 8 12 25

B 10 6 4 20

C 8 2 5 15

23 16 21 60

a) P(A D) = P(A) + P(D) - P(A_ D) = 25/60 + 23/60 - 5/60 = 43/60 = .7167

b) P(E B) = P(E) + P(B) - P(E_B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000

c) P(D E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500

d) P(C F) = P(C) + P(F) - P(C_ F) = 15/60 + 21/60 - 5/60 = 31/60 = .5167


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4.10

E F

A .10 .03 .13

B .04 .12 .16

C .27 .06 .33

D .31 .07 .38

.72 .28 1.00

a) P(A F) = P(A) + P(F) - P(A_ F) = .13 + .28 - .03 = .38b) P(E B) = P(E) + P(B) - P(E_B) = .72 + .16 - .04 = .84c) P(B C) = P(B) + P(C) =.16 + .33 = .49d) P(E F) = P(E) + P(F) = .72 + .28 = 1.000

4.11 A = event - flown in an airplane at least once

T = event - ridden in a train at least once

P(A) = .47 P(T) = .28

P (ridden either a train or an airplane) =

P(A T) = P(A) + P(T) - P(A_ T) = .47 + .28 - P(A_ T)

Cannot solve this problem without knowing the probability of the intersection.

We need to know the probability of the intersection of A and T, the proportion

who have ridden both.

4.12 P(L) = .75 P(M) = .78 P(M L) = .61

a) P(M L) = P(M) + P(L) - P(M_ L) = .78 + .75 - .61 = .92b) P(M L) but not both = P(M L) - P(M_ L) = .92 - .61 = .31c) P(NM_ NL) = 1 - P(M L) = 1 - .92 = .08

4.13 Let C = have cable TV

Let T = have 2 or more TV sets

P(C) = .67, P(T) = .74, P(C T) = .55

a) P(C T) = P(C) + P(T) - P(C_ T) = .67 + .74 - .55 = .86b) P(C T but not both) = P(C T) - P(C_T) = .86 - .55 = .31


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c) P(NC_ NT) = 1 - P(C T) = 1 - .86 = .14d) The special law of addition does not apply because P(C _ T) is not .0000.

Possession of cable TV and 2 or more TV sets are not mutually exclusive.

4.14 Let T = review transcript

F = consider faculty references

P(T) = .54 P(F) = .44 P(T_ F) = .35

a) P(F T) = P(F) + P(T) - P(F_ T) = .44 + .54 - .35 = .63b) P(F T) - P(F_ T) = .63 - .35 = .28c) 1 - P(F T) = 1 - .63 = .37

d)

Y N

Y .35 .19 .54

N .09 .37 .46

.44 .56 1.00

4.15

C D E F

A 5 11 16 8 40

B 2 3 5 7 17

7 14 21 15 57

a) P(A_ E) = 16/57 = .2807b) P(D_ B) = 3/57 = .0526c) P(D_ E) = .0000d) P(A_ B) = .0000

4.16

D E F

A .12 .13 .08 .33

B .18 .09 .04 .31

C .06 .24 .06 .36


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.36 .46 .18 1.00

a) P(E_ B) = .09b) P(C_F) = .06c) P(E_D) = .00

4.17 Let D = Defective part

a) (without replacement)

P(D1_ D2) = P(D1) P(D2D1) =2450

30

49

5

50

6= = .0122

b) (with replacement)

P(D1_ D2) = P(D1) P(D2) =2500

36

50

6

50

6= = .0144

4.18 Let U = Urban

I = care for Ill relatives

a) P(U_ I) = P(U) P(I U)

P(U) = .78 P(I) = .15 P(I

U) = .11

P(U_ I) = (.78)(.11) = .0858

b) P(U_NI) = P(U) P(NIU) but P(IU) = .11So, P(NIU) = 1 - .11 = .89 and P(U_NI) =

P(U) P(NIU) = (.78)(.89) = .6942

c)

U

Yes No

I

Yes .15

No .85

.78 .22

The answer to a) is found in the YES-YES cell. To compute this cell, take 11%


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or .11 of the total (.78) people in urban areas. (.11)(.78) = .0858 which belongs in

the YES-YES" cell. The answer to b) is found in the Yes for U and no for I cell.It can be determined by taking the marginal, .78, less the answer for a), .0858.

d. P(NU_ I) is found in the no for U column and the yes for I row (1strow and2

ndcolumn). Take the marginal, .15, minus the yes-yes cell, .0858, to get

.0642.

4.19 Let S = stockholder

Let C = college

P(S) = .43 P(C) = .37 P(CS) = .75

a) P(NS) = 1 - .43 = .57

b) P(S_C) = P(S)P(CS) = (.43)(.75) = .3225c) P(S C) = P(S) + P(C) - P(S_C) = .43 + .37 - .3225 = .4775d) P(NS_NC) = 1 - P(S C) = 1 - .4775 = .5225e) P(NS NC) = P(NS) + P(NC) - P(NS_NC) = .57 + .63 - .5225 = .6775f) P(C_ NS) = P(C) - P(C_ S) = .37 - .3225 = .0475

4.20 Let F = fax machine

Let P = personal computer

Given: P(F) = .10 P(P) = .52 P(PF) = .91

a) P(F_ P) = P(F) P(PF) = (.10)(.91) = .091

b) P(F P) = P(F) + P(P) - P(F_ P) = .10 + .52 - .091 = .529

c) P(F_ NP) = P(F) P(NPF)

Since P(PF) = .91, P(NPF)= 1 - P(PF) = 1 - .91 = .09P(F_ NP) = (.10)(.09) = .009

d) P(NF_ NP) = 1 - P(F P) = 1 - .529 = .471e) P(NF_ P) = P(P) - P(F_ P) = .52 - .091 = .429

P NP

F .091 .009 .10

NF .429 .471 .90

.520 .480 1.00

4.21Let S = safety


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Let A = age

P(S) = .30 P(A) = .39 P(AS) = .87

a) P(S_ NA) = P(S)P(NAS)

but P(NAS) = 1 - P(AS) = 1 - .87 = .13P(S_ NA) = (.30)(.13) = .039

b) P(NS_ NA) = 1 - P(S A) = 1 - [P(S) + P(A) - P(S_A)]but P(S_A) = P(S) P(AS) = (.30)(.87) = .261P(NS_ NA) = 1 - (.30 + .39 - .261) = .571

c) P(NS_A) = P(NS) - P(NS_NA)

but P(NS) = 1 - P(S) = 1 - .30 = .70

P(NS_A) = .70 - 571 = .129

4.22 Let C = ceiling fans

Let O = outdoor grill

P(C) = .60 P(O) = .29 P(C O) = .13

a) P(C O)= P(C) + P(O) - P(C_ O) = .60 + .29 - .13 = .76

b) P(NC_ NO) = 1 - P(C O)= 1 - .76 = .24

c) P(NC_ O) = P(O) - P(C_ O) = .29 - .13 = .16

d) P(C_ NO) = P(C) - P(C_ O) = .60 - .13 = .47

4.23

E F G

A 15 12 8 35

B 11 17 19 47

C 21 32 27 80

D 18 13 12 43

65 74 66 205

a) P(GA) = 8/35 = .2286

b) P(BF) = 17/74 = .2297


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c) P(CE) = 21/65 = .3231

d) P(EG) = .0000

4.24

C D

A .36 .44 .80

B .11 .09 .20

.47 .53 1.00

a) P(CA) = .36/.80 = .4500

b) P(BD) = .09/.53 = .1698

c) P(AB) = .0000

4.25

Calculator

Yes No

ComputerYes 46 3 49

No 11 15 26

57 18 75

Select a category from each variable and test

P(V1V2) = P(V1).

For example, P(Yes ComputerYes Calculator) = P(Yes Computer)?

75

49

57

46= ?

.8070 .6533

Variable of Computer not independent of Variable of Calculator.


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4.26

Let C = constructionLet S = South Atlantic

83,384 total failures

10,867 failures in construction

8,010 failures in South Atlantic

1,258 failures in construction and South Atlantic

a) P(S) = 8,010/83,384 = .09606

b) P(C S) = P(C) + P(S) - P(C_ S) =

10,867/83,384 + 8,010/83,384 - 1,258/83,384 = 17,619/83,384 = .2113

c) P(C S) =

384,83

8010

384,83

1258

)(

)(=

SP

SCP= .15705

d) P(S C) =

384,83

867,10

384,83

1258

)(

)(=

CP

SCP = .11576

e) P(NSNC) =)(

)(1

)(

)(

NCP

SCP

NCP

NCNSP =

but NC = 83,384 - 10,867 = 72,517

and P(NC) = 72,517/83,384 = .869675

Therefore, P(NSNC) = (1 - .2113)/(.869675) = .9069

f) P(NS C) =)(

)()(

)(

)(

CP

SCPCP

CP

CNSP =

but P(C) = 10,867/83,384 = .1303

P(C_ S) = 1,258/83,384 = .0151

Therefore, P(NSC) = (.1303 - .0151)/.1303 = .8842

4.27 Let E = Economy

Let Q = Qualified


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P(E) = .46 P(Q) = .37 P(E_Q) = .15

a) P(EQ) = P(E_Q)/P(Q) = .15/.37 = .4054

b) P(QE) = P(E_Q)/P(E) = .15/.46 = .3261

c) P(QNE) = P(Q_NE)/P(NE)

but P(Q_ NE) = P(Q) - P(Q_E) = .37 - .15 = .22

P(NE) = 1 - P(E) = 1 - .46 = .54

P(QNE) = .22/.54 = .4074

d) P(NE_NQ) = 1 - P(E Q) = 1 - [P(E) + P(Q) - P(E _Q)]

= 1 - [.46 + .37 + .15] = 1 - (.68) = .32

4.28

Let A = airline tickets

Let T = transacting loans

P(A) = .47 P(TA) = .81

a) P(A_T) = P(A) P(TA) = (.47)(.81) = .3807b) P(NTA) = 1 - P(TA) = 1 - .81 = .19

c) P(NT_ A) = P(A) - P(A_ T) = .47 - .3807 = .0893

4.29 Let H = hardware

Let S = software

P(H) = .37 P(S) = .54 P(SH) = .97

a) P(NSH) = 1 - P(SH) = 1 - .97 = .03

b) P(SNH) = P(S_NH)/P(NH)but P(H_ S) = P(H) P(SH) = (.37)(.97) = .3589so P(NH_ S) = P(S) - P(H_ S) = .54 - .3589 = .1811

P(NH) = 1 - P(H) = 1 - .37 = .63

P(SNH) = (.1811)/(.63) = .2875

c) P(NHS) = P(NH_S)/P(S) = .1811//54 = .3354


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d) P(NHNS) = P(NH_NS)/P(NS)but P(NH_NS) = P(NH) - P(NH_ S) = .63 - .1811 = .4489

and P(NS) = 1 - P(S) = 1 - .54 = .46

P(NHNS) = .4489/.46 = .9759

4.30 Let A = product produced on Machine AB = product produces on Machine B

C = product produced on Machine C

D = defective product

P(A) = .10 P(B) = .40 P(C) = .50

P(DA) = .05 P(DB) = .12 P(DC) = .08

Event Prior Conditional Joint Revised

P(Ei) P(DEi)P(D_Ei)

A .10 .05 .005 .005/.093=.0538

B .40 .12 .048 .048/.093=.5161

C .50 .08 .040 .040/.093=.4301

P(D)=.093

Revise:P(AD) = .005/.093= .0538

P(BD) = .048/.093 = .5161

P(CD) = .040/.093 = .4301

4.31 Let A = Alex fills the order

B = Alicia fills the order

C = Juan fills the order

I = order filled incorrectly

K = order filled correctly

P(A) = .30 P(B) = .45 P(C) = .25

P(IA) = .20 P(IB) = .12 P(IC) = .05P(KA) = .80 P(KB) = .88 P(KC) = .95

a) P(B) = .45

b) P(KC) = 1 - P(IC) = 1 - .05 = .95


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c)


P(Ei) P(IEi)P(I_Ei)

P(EiI)A .30 .20 .0600 .0600/.1265=.4743

B .45 .12 .0540 .0540/.1265=.4269

C .25 .05 .0125 .0125/.1265=.0988

P(I)=.1265

Revised: P(AI) = .0600/.1265 = .4743

P(BI) = .0540/.1265 = .4269

P(CI) = .0125/.1265 = .0988

d)


P(Ei) P(KEi)P(K_Ei)

P(EiK)A .30 .80 .2400 .2400/.8735=.2748

B .45 .88 .3960 .3960/.8735=.4533

C .25 .95 .2375 .2375/.8735=.2719P(K)=.8735

4.32 Let T = lawn treated by Tri-state

G = lawn treated by Green Chem

V = very healthy lawn

N = not very healthy lawn

P(T) = .72 P(G) = .28 P(VT) = .30 P(VG) = .20


P(Ei) P(VEi)P(V_Ei)

P(EiV)A .72 .30 .216 .216/.272=.7941

B .28 .20 .056 .056/.272=.2059

P(V)=.272

Revised: P(TV) = .216/.272 = .7941

P(GV) = .056/.272 = .2059


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4.33 Let T = training

Let S = small

P(T) = .65

P(ST) = .18 P(NST) = .82P(SNT) = .75 P(NSNT) = .25


P(Ei) P(NSEi)P(NS_Ei)

P(EiNS)T .65 .82 .5330 .5330/.6205=.8590

NT .35 .25 .0875 .0875/.6205=.1410

P(NS)=.6205

4.34

Variable 1

D E

A 10 20

Variable 2 B 15 5

C 30 15

55 40 95

a) P(E) = 40/95 = .42105

b) P(B D) = P(B) + P(D) - P(B_D)

= 20/95 + 55/95 - 15/95 = 60/95 = .63158

c) P(A_E) = 20/95 = .21053

d) P(BE) = 5/40 = .1250e) P(A B) = P(A) + P(B) = 30/95 + 20/95 =

50/95 = .52632

f) P(B_C) = .0000 (mutually exclusive)

g) P(DC) = 30/45 = .66667

h) P(AB)= P(A_B) = .0000 = .0000 mutually exclusiveP(B) 20/95


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i) P(A) = P(AD)??

30/95 = 10/95 ??

.31579 .18182No, Variables 1 and 2 are not independent.

4.35

D E F G

A 3 9 7 12 31

B 8 4 6 4 22

C 10 5 3 7 25

21 18 16 23 78

a) P(F_A) = 7/78 = .08974b) P(AB) = P(A_B) = .0000 = .0000

P(B) 22/78

c) P(B) = 22/78 = .28205

d) P(E_F) = .0000 Mutually Exclusive

e) P(DB) = 8/22 = .36364

f) P(BD) = 8/21 = .38095g) P(D C) = 21/78 + 25/78 10/78 = 36/78 = .4615

h) P(F) = 16/78 = .20513

4.36

Age(years)

65

Gender Male .11 .20 .19 .12 .16 .78

Female .07 .08 .04 .02 .01 .22

.18 .28 .23 .14 .17 1.00

a) P(35-44) = .28

b) P(Woman_45-54) = .04


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c) P(Man 35-44) = P(Man) + P(35-44) - P(Man_35-44) = .78 + .28 - .20 = .86d) P(


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P(NT) = 1 - P(T) = 1 - .16 = .84

Therefore, P(NENT) = {P(NE)P(NTNE)}/P(NT) =

{(.20)(.83)}/(.84) = .1976

e) P(not W_not NET) = P(not W_not NE_T)/ P(T)but P(not W_not NE_T) =.16 - P(W_T) - P(NE_T) = .16 - .042 - .034 = .084P(not W_not NE_T)/ P(T) = (.084)/(.16) = .525

4.40 Let M = Mastercard A = American Express V = Visa

P(M) = .30 P(A) = .20 P(V) = .25

P(M_A) = .08 P(V_M) = .12 P(A_V) = .06

a) P(V A) = P(V) + P(A) - P(V_A)

= .25 + .20 - .06 = .39

b) P(VM) = P(V_M)/P(M) = .12/.30 = .40c) P(MV) = P(V_M)/P(V) = .12/.25 = .48

d) P(V) = P(VM)??

.25 .40

Possession of Visa is not independent of

possession of Mastercard

e) American Express is not mutually exclusive of Visa

because P(A_V) .0000

4.41

Let S = believe SS secure

N = don't believe SS will be secure

45 = 45 or more years old

P(N) = .51

Therefore, P(S) = 1 - .51 = .49

P(S>45) = .70Therefore, P(N>45) = 1 - P(S>45) = 1 - .70 = .30

P(


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a) P(>45) = 1 - P(45_S) = P(>45)P(S>45) = (.57)(.70) = .301P(45_S) = .49 - .301 = .189

c) P(>45S) = P(>45_S)/P(S) = P(>45)P(S>45)/P(S) =

(.43)(.70)/.49 = .6143

d) (


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SaveYes .3483 .0817 .43

No .1017 .4683 .57

.4500 .5500 1.00

4.43 Let R = readLet B = checked in the with boss

P(R) = .40 P(B) = .34 P(BR) = .78

a) P(B_R) = P(R)P(BR) = (.40)(.78) = .312b) P(NR_NB) = 1 - P(R B)

but P(R B) = P(R) + P(B) - P(R_B) =

.40 +.34 - .312 = .428

P(NR_NB) = 1 - .428 = .572c) P(RB) = P(R_B)/P(B) = (.312)/(.34) = .9176

d) P(NBR) = 1 - P(BR) = 1 - .78 = .22e) P(NBNR) = P(NB_NR)/P(NR)

but P(NR) = 1 - P(R) = 1 - .40 = .60

P(NBNR) = .572/.60 = .9533

f) Probability matrix for problem 4.43:

B NB

R .312 .088 .40

NR .028 .572 .60

.340 .660 1.00

4.44

Let: D = denial

I = inappropriate

C = customer

P = payment dispute

S = specialtyG = delays getting care


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R = prescription drugs

P(D) = .17 P(I) = .14 P(C) = .14 P(P) = .11

P(S) = .10 P(G) = .08 P(R) = .07

a) P(P S) = P(P) + P(S) = .11 + .10 = .21b) P(R_C) = .0000 (mutually exclusive)c) P(IS) = P(I_S)/P(S) = .0000/.10 = .0000d) P(NG_NP) = 1 - P(G P) = 1 - [P(G) + P(P)] =

1 [.08 + .11] = 1 - .19 = .81

4.45

Let R = retention

P = process

P(R) = .56 P(P_R) = .36 P(RP) = .90

a) P(R_NP) = P(R) - P(P_R) = .56 - .36 = .20b) P(PR) = P(P_R)/P(R) = .36/.56 = .6429

c) P(P) = ??

P(RP) = P(R_P)/P(P)so P(P) = P(R_P)/P(RP) = .36/.90 = .40

d) P(R P) = P(R) + P(P) - P(R_P) =

.56 + .40 - .36 = .60

e) P(NR_NP) = 1 - P(R P) = 1 - .60 = .40f) P(RNP) = P(R_NP)/P(NP)

but P(NP) = 1 - P(P) = 1 - .40 = .60

P(RNP) = .20/.60 = .33

4.46 Let M = mail

S = sales

P(M) = .38 P(M_S) = .0000 P(NM_NS) = .41

a) P(M_NS) = P(M) - P(M_S) = .38 - .00 = .38

b) P(S):

P(M S) = 1 - P(NM_NS) = 1 - .41 = .59P(M S) = P(M) + P(S)Therefore, P(S) = P(M S) - P(M) = .59 - .38 = .21

c) P(SM) = P(S_M)/P(M) = .00/.38 = .00


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d) P(NMNS) = P(NM_NS)/P(NS) = .41/.79 = .5190

where: P(NS) = 1 - P(S) = 1 - .21 = .79

4.47 Let S = Sarabia

T = Tran

J = JacksonB = blood test

P(S) = .41 P(T) = .32 P(J) = .27

P(B S) = .05 P(B T) = .08 P(B J) = .06


P(Ei) P(BEi)P(B_Ei)

P(BiNS)S .41 .05 .0205 .329

T .32 .08 .0256 .411

J .27 .06 .0162 .260

P(B)=.0623

4.48 Let R = regulationsT = tax burden

P(R) = .30 P(T) = .35 P(TR) = .71

a) P(R_T) = P(R)P(TR) = (.30)(.71) = .2130b) P(R T) = P(R) + P(T) - P(R_T) =

.30 + .35 - .2130 = .4370

c) P(R T) - P(R_T) = .4370 - .2130 = .2240d) P(RT) = P(R_T)/P(T) = .2130/.35 = .6086

e) P(NRT) = 1 - P(RT) = 1 - .6086 = .3914f) P(NRNT) = P(NR_NT)/P(NT) = [1 - P(R T)]/P(NT) =

(1 - .4370)/.65 = .8662

4.49


P(Ei) P(NSEi)P(NS_Ei)

P(EiNS)Soup .60 .73 .4380 .8456

Breakfast


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Meats .35 .17 .0595 .1149

Hot Dogs .05 .41 .0205 .0396

.5180

4.50 Let GH = Good health

HM = Happy marriage

FG = Faith in God

P(GH) = .29 P(HM) = .21 P(FG) = .40

a) P(HM FG) = P(HM) + P(FG) - P(HM_FG)but P(HM_FG) = .0000

P(HM FG) = P(HM) + P(FG) = .21 + .40 = .61b) P(HM FG GH) = P(HM) + P(FG) + P(GH) =

.29 + .21 + .40 = .9000

c) P(FG_GH) = .0000

The categories are mutually exclusive.

The respondent could not select more than one

answer.

d) P(neither FG nor GH nor HM) =

1 - P(HM FG GH) = 1 - .9000 = .1000

Chapter 4 Managerial Statistics solutions

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